Straight Lines
Area and Locus
Grade 11
Question:
<p>Let A(1, 2), B(3, 4) be two points and C(x, y) be a point such that \((x-1)(x-3) + (y-2)(y-4) = 0\). If area of \(\triangle ABC\) is 1 sq unit, then the maximum number of positions of C in the XY-plane, is</p>
<p>(a) 2</p>
<p>(b) 4</p>
<p>(c) 8</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The constraint defines a circle, and points on this circle satisfying the area condition form discrete positions. Geometric analysis shows there are at most 4 such positions.
<p><strong>Solution:</strong></p><p>The constraint $(x-1)(x-3) + (y-2)(y-4) = 0$ can be rewritten as:</p><p>$$x^2 - 4x + 3 + y^2 - 6y + 8 = 0$$</p><p>$$(x-2)^2 - 1 + (y-3)^2 - 1 = 0$$</p><p>$$(x-2)^2 + (y-3)^2 = 2$$</p><p>This is a circle with center $M(2, 3)$ and radius $\sqrt{2}$.</p><p>Point C lies on this circle, and the area of $\triangle ABC = 1$ sq unit.</p><p>The distance from C to line AB determines the area. Since the circle has a fixed radius and center, there can be at most 4 positions where C satisfies both the circle equation and the area constraint (2 on each side of line AB where the perpendicular distance from the circle to AB equals the required distance).</p><p>∴ Answer is (b)</p>
Correct Answer: B